Green’s Theorem

Cover Page

Artifacts

Artifact #1	Line Integral by Green’s Theorem (HW19 #3)
Artifact #2	Area of a Parametric Region by Green’s Theorem (HW19 #4)

Sub-outcome coverage

Sub Outcome	Artifact #1	Artifact #2
Do a double integral instead: calculate a line integral by using Green’s Theorem to evaluate an equivalent double integral.	Yes
Do a line integral instead: calculate a double integral by using Green’s Theorem to evaluate an equivalent line integral.		Yes
Area of a region: use Green’s Theorem to calculate the area of a two-dimensional region bounded by lines or curves.		Yes

Artifact 1

Problem

Use Green’s Theorem to evaluate \[ \oint_{C} \vec{F} \cdot d\vec{r}, \qquad \vec{F} = \left\langle \sqrt{x} + 5y, \; 2x + \sqrt{y} \right\rangle, \] where \(C\) consists of the arc of the curve \(y = x - x^{2}\) from \((0, 0)\) to \((1, 0)\) and the line segment from \((1, 0)\) back to \((0, 0)\).

Solution:

The two square roots make a direct line integral unpleasant, so we trade the line integral for a double integral with Green’s Theorem.

Green’s Theorem

For a positively oriented (counterclockwise) simple closed curve \(C\) bounding a region \(D\), with \(\vec{F} = \langle P, Q \rangle\), \[ \oint_{C} P\, dx + Q\, dy = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA. \]

First we check the orientation. The arc \(y = x - x^{2}\) is the top boundary and the segment along the \(x\)-axis is the bottom boundary. Traveling along the arc from \((0, 0)\) to \((1, 0)\) and then back along the axis from \((1, 0)\) to \((0, 0)\) runs clockwise, which is the negative orientation. Green’s Theorem is stated for the counterclockwise direction, so we attach a minus sign: \[ \oint_{C} \vec{F} \cdot d\vec{r} = -\iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA. \]

Interactive

The shaded region is \(D\). Drag the slider to send a point around \(C\) in the order the problem gives, first along the arc and then along the segment, and watch that it travels clockwise.

viewof tpos = Inputs.range([0, 1], {step: 0.01, value: 0.25, label: "position along C"})

boundaryPt = {
  if (tpos <= 0.5) {
    const s = tpos / 0.5;          // arc (0,0) -> (1,0)
    return [s, s - s * s];
  } else {
    const s = (tpos - 0.5) / 0.5;  // segment (1,0) -> (0,0)
    return [1 - s, 0];
  }
}

arcPts = Array.from({length: 81}, (_, i) => { const x = i / 80; return [x, x - x * x]; })

Plot.plot({
  width: 460,
  height: 320,
  marginLeft: 45,
  style: {background: "transparent", color: "#334155"},
  x: {label: "x", grid: true, domain: [-0.1, 1.1]},
  y: {label: "y", grid: true, domain: [-0.1, 0.32]},
  marks: [
    Plot.areaY(arcPts, {x: d => d[0], y: d => d[1], fill: "#cfe3f7", fillOpacity: 0.7}),
    Plot.line(arcPts, {x: d => d[0], y: d => d[1], stroke: "#2563eb", strokeWidth: 2}),
    Plot.line([[1, 0], [0, 0]], {x: d => d[0], y: d => d[1], stroke: "#2563eb", strokeWidth: 2}),
    Plot.dot([boundaryPt], {x: d => d[0], y: d => d[1], fill: "#d97706", r: 7})
  ]
})

Now we read off the partial derivatives. With \(P = \sqrt{x} + 5y\) and \(Q = 2x + \sqrt{y}\), \[ \frac{\partial Q}{\partial x} = 2, \qquad \frac{\partial P}{\partial y} = 5, \qquad \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2 - 5 = -3. \]

The square roots are gone, and the integrand is a constant. The region is \[ D = \left\{ (x, y) : 0 \le x \le 1, \; 0 \le y \le x - x^{2} \right\}, \] so \[ \begin{aligned} \oint_{C} \vec{F} \cdot d\vec{r} &= -\iint_{D} (-3)\, dA \\[4pt] &= 3 \iint_{D} dA \\[4pt] &= 3 \int_{0}^{1} \int_{0}^{\,x - x^{2}} 1 \, dy \, dx \\[4pt] &= 3 \int_{0}^{1} (x - x^{2}) \, dx \\[4pt] &= 3 \left[ \frac{x^{2}}{2} - \frac{x^{3}}{3} \right]_{0}^{1} \\[4pt] &= 3 \left( \frac{1}{2} - \frac{1}{3} \right) \\[4pt] &= 3 \cdot \frac{1}{6} \\[4pt] &= \frac{1}{2}. \end{aligned} \]

Artifact 2

Problem

Find the area of the region enclosed by \[ x = t - t^{5}, \qquad y = t - t^{7}, \qquad 0 \le t \le 1, \] using Green’s Theorem.

Solution:

We start with the definition of area as a double integral over the enclosed region \(D\): \[ A = \iint_{D} 1 \, dA. \]

Setting this up directly is the hard part. The boundary is given parametrically, and to write \(D\) as \(0 \le x \le x_{\max}\), \(g(x) \le y \le h(x)\) we would need \(y\) as a function of \(x\). Eliminating \(t\) from \(x = t - t^{5}\) is not possible in any closed form, so there is no clean pair of \(y\)-limits to integrate between.

Area from Green’s Theorem

Taking \(P = -\tfrac{1}{2} y\) and \(Q = \tfrac{1}{2} x\) gives \(\dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y} = \tfrac{1}{2} + \tfrac{1}{2} = 1\), so \[ A = \iint_{D} 1 \, dA = \frac{1}{2} \oint_{C} \left( x \, dy - y \, dx \right). \]

Before applying it we confirm \(C\) is a closed curve. At \(t = 0\) we get \((x, y) = (0, 0)\), and at \(t = 1\) we get \((1 - 1, \, 1 - 1) = (0, 0)\), so the curve starts and ends at the same point.

Interactive

Drag the slider to trace the curve as \(t\) runs from \(0\) to \(1\). The orange dot is the current point, the blue curve is the path traced so far, and the shaded region is the area we are computing.

viewof tparam = Inputs.range([0, 1], {step: 0.01, value: 0.5, label: "t"})

fullCurve = Array.from({length: 121}, (_, i) => { const t = i / 120; return [t - t ** 5, t - t ** 7]; })

tracedCurve = fullCurve.filter((_, i) => i / 120 <= tparam + 1e-9)

curPt = [tparam - tparam ** 5, tparam - tparam ** 7]

Plot.plot({
  width: 460,
  height: 340,
  marginLeft: 45,
  style: {background: "transparent", color: "#334155"},
  x: {label: "x", grid: true},
  y: {label: "y", grid: true},
  marks: [
    Plot.areaY(fullCurve, {x: d => d[0], y: d => d[1], fill: "#cfe3f7", fillOpacity: 0.6}),
    Plot.line(tracedCurve, {x: d => d[0], y: d => d[1], stroke: "#2563eb", strokeWidth: 2.5}),
    Plot.dot([curPt], {x: d => d[0], y: d => d[1], fill: "#d97706", r: 7})
  ]
})

Now we set up the line integral from the parametrization. The differentials are \[ dx = \left( 1 - 5t^{4} \right) dt, \qquad dy = \left( 1 - 7t^{6} \right) dt. \]

Form the integrand \(x \, dy - y \, dx\): \[ \begin{aligned} x \, dy - y \, dx &= \left( t - t^{5} \right)\!\left( 1 - 7t^{6} \right) dt - \left( t - t^{7} \right)\!\left( 1 - 5t^{4} \right) dt \\[4pt] &= \left( t - 7t^{7} - t^{5} + 7t^{11} \right) dt - \left( t - 5t^{5} - t^{7} + 5t^{11} \right) dt \\[4pt] &= \left( 4t^{5} - 6t^{7} + 2t^{11} \right) dt. \end{aligned} \]

So the area is \[ \begin{aligned} A = \frac{1}{2} \oint_{C} \left( x \, dy - y \, dx \right) &= \frac{1}{2} \int_{0}^{1} \left( 4t^{5} - 6t^{7} + 2t^{11} \right) dt \\[4pt] &= \frac{1}{2} \left[ \frac{4t^{6}}{6} - \frac{6t^{8}}{8} + \frac{2t^{12}}{12} \right]_{0}^{1} \\[4pt] &= \frac{1}{2} \left( \frac{2}{3} - \frac{3}{4} + \frac{1}{6} \right) \\[4pt] &= \frac{1}{2} \cdot \frac{8 - 9 + 2}{12} \\[4pt] &= \frac{1}{2} \cdot \frac{1}{12} \\[4pt] &= \frac{1}{24}. \end{aligned} \]

Essay

Your general approach to problems like this:

My general approach to a Green’s Theorem problem is to first decide which direction across the theorem saves the most work, because the equality lets me trade a line integral for a double integral or trade a double integral for a line integral. I look at which side of the problem is the awkward one and then convert to the other side. In the first artifact the field carried two square roots, which are unpleasant to integrate along a curve, so trading the line integral for a double integral was the better direction. In the second artifact the region was pinned down only by a parametric curve I could not solve for \(y\), so trading the double integral for a line integral was the better direction.

An evaluation of your strengths and areas for improvement in this section:

My strength in this section is reading the problem for that decision and checking orientation before I commit, since the first curve ran clockwise and the wrong sign would have flipped the answer. The area I could use improvement on is the bookkeeping once the setup is done. Tracking signs through the orientation check and keeping the parametric algebra straight are the steps where a small slip would change the result, and those are exactly the kind of mistakes I make when I move too fast.

Describe how the artifacts you provided support your self-evaluation:

My self-evaluation is supported by both artifacts. In the first artifact the combination \(\partial Q / \partial x - \partial P / \partial y\) collapsed to the constant \(-3\), so the work was in correctly recognizing the clockwise orientation and attaching the minus sign rather than in the integration. In the second artifact the conversion to the line integral \(\tfrac{1}{2}\oint (x\, dy - y\, dx)\) turned an impossible setup into a routine polynomial integral, where the only real risk was an algebra slip while expanding and combining the parametric terms.

What are the critical pieces of a problem of this type?

I would list them in the following order: choosing which direction across the theorem to apply, checking the orientation of the curve and attaching the correct sign, and then carrying out the setup and computation accurately.